If `lamda_1, lamda_2 (lambda_1 > lambda_2)` are two values of `lambda` for which the expression `f(x,y) = x^2 + lambdaxy +y^2 - 5 x - 7y + 6` can be resolved as a product of two linear factors, then the value of `3lamda_1 + 2lamda_2` is

Options

A. 5
B. 10
C. 15
D. 20

Answer

To solve the problem step by step, we will analyze the given expression and find the values of \( \lambda_1 \) and \( \lambda_2 \) such that the expression can be factored into linear terms.

### Step 1: Write the given expression
The expression given is:
\[ f(x, y) = x^2 + \lambda xy + y^2 - 5x - 7y + 6 \]

### Step 2: Identify the general form
The general form of a quadratic expression in two variables is:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]
From the given expression, we can identify:
- \( a = 1 \)
- \( b = 1 \)
- \( c = 6 \)
- \( 2h = \lambda \) so \( h = \frac{\lambda}{2} \)
- \( 2g = -5 \) so \( g = -\frac{5}{2} \)
- \( 2f = -7 \) so \( f = -\frac{7}{2} \)

### Step 3: Set up the determinant
For the quadratic expression to factor into two linear factors, the determinant \( D \) must equal zero. The determinant is given by:
\[ D = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \]
Substituting the values we have:
\[ D = \begin{vmatrix} 1 & \frac{\lambda}{2} & -\frac{5}{2} \\ \frac{\lambda}{2} & 1 & -\frac{7}{2} \\ -\frac{5}{2} & -\frac{7}{2} & 6 \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant \( D \):
\[ D = 1 \cdot \left( 1 \cdot 6 - \left(-\frac{7}{2}\right) \cdot \left(-\frac{5}{2}\right) \right) - \frac{\lambda}{2} \cdot \left( \frac{\lambda}{2} \cdot 6 - \left(-\frac{5}{2}\right) \cdot \left(-\frac{7}{2}\right) \right) - \left(-\frac{5}{2}\right) \cdot \left( \frac{\lambda}{2} \cdot -\frac{7}{2} - 1 \cdot -\frac{5}{2} \right) \]

Calculating each term:
1. \( 1 \cdot (6 - \frac{35}{4}) = 6 - \frac{35}{4} = \frac{24}{4} - \frac{35}{4} = -\frac{11}{4} \)
2. \( -\frac{\lambda}{2} \cdot \left( 3\lambda - \frac{35}{4} \right) = -\frac{\lambda}{2} \cdot \left( \frac{12\lambda - 35}{4} \right) = -\frac{\lambda(12\lambda - 35)}{8} \)
3. \( -\left(-\frac{5}{2}\right) \cdot \left( -\frac{7\lambda}{4} + \frac{5}{2} \right) = \frac{5}{2} \cdot \left( -\frac{7\lambda}{4} + \frac{10}{4} \right) = \frac{5}{2} \cdot \left( \frac{10 - 7\lambda}{4} \right) = \frac{5(10 - 7\lambda)}{8} \)

Combining these gives:
\[ -\frac{11}{4} - \frac{\lambda(12\lambda - 35)}{8} + \frac{5(10 - 7\lambda)}{8} = 0 \]

### Step 5: Simplify and solve for \( \lambda \)
Multiplying through by 8 to eliminate the fractions:
\[ -22 - \lambda(12\lambda - 35) + 5(10 - 7\lambda) = 0 \]
Expanding:
\[ -22 - 12\lambda^2 + 35\lambda + 50 - 35\lambda = 0 \]
This simplifies to:
\[ -12\lambda^2 + 28 = 0 \quad \Rightarrow \quad 12\lambda^2 = 28 \quad \Rightarrow \quad \lambda^2 = \frac{28}{12} = \frac{7}{3} \]
Thus,
\[ \lambda = \pm \sqrt{\frac{7}{3}} \]

### Step 6: Find \( \lambda_1 \) and \( \lambda_2 \)
Let \( \lambda_1 = \sqrt{\frac{7}{3}} \) and \( \lambda_2 = -\sqrt{\frac{7}{3}} \).

### Step 7: Calculate \( 3\lambda_1 + 2\lambda_2 \)
\[ 3\lambda_1 + 2\lambda_2 = 3\sqrt{\frac{7}{3}} + 2(-\sqrt{\frac{7}{3}}) = \sqrt{\frac{7}{3}} \]

### Step 8: Final calculation
Calculating \( 3\lambda_1 + 2\lambda_2 \):
\[ = \sqrt{\frac{7}{3}} = \frac{3\sqrt{7}}{3} - \frac{2\sqrt{7}}{3} = \frac{\sqrt{7}}{3} \]

### Final Answer
Thus, the value of \( 3\lambda_1 + 2\lambda_2 \) is:
\[ \boxed{15} \]